Problem: $\overline{AC}$ is $6$ units long $\overline{BC}$ is $8$ units long $\overline{AB}$ is $10$ units long What is $\sec(\angle BAC)?$ A C B 6 8 10
Solution: $\sec(\angle BAC) = \dfrac{1}{\cos(\angle BAC)}$ How can we find $\cos(\angle BAC)$ SOH CAH TOA osine = djacent over ypotenuse Adjacent $= \overline{AC} = 6$ Hypotenuse $= \overline{AB} = 10$ $\cos(\angle BAC) = \dfrac{6}{10}$ $\sec(\angle BAC) = \dfrac{1}{\cos(\angle BAC)} = \dfrac{10}{6}$